(a) A pressure vessel contains a large volume of CO2 gas at 10 atm pressure. A membrane composed of a poly(ether ketone) polymer with thickness 100 microns and net effective area of 100 cm2 covers a small perforated port in the container. The solubility of CO2 at 10 atm is 6.97 x 10-4 moles/cm3 at 35 ∘C. The diffusivity of CO2 in the polymer is known to be 2.29 x 10-8 cm2/s at this temperature. How long will it take for 0.001 moles of CO2 to leak from the container at steady-state? Assume that the amount of carbon dioxide in the surroundings is insignificant. Express your answer in hours.
(b) If the diffusivity is observed to double when the temperature is increased by 10 degrees C, what is the activation energy for diffusion? Express your answer in units of eV.
a) 1.74
a) it's right
pls what about b
39.2
Is 39.2 in eV?
Is 39.2 in KJ/mol? please help
its not right with 39.2 and after conversion with 1KJ/mol=1.03*10^-2EV. Any ideas or thoughts on b)?
0.551
(a) To find the time it takes for 0.001 moles of CO2 to leak from the container at steady-state, we can use Fick's Law of diffusion:
J = -D * dC/dx
Where:
- J is the molar flux (moles/(cm^2.s))
- D is the diffusivity of CO2 in the polymer (cm^2/s)
- dC/dx is the concentration gradient (moles/cm^4)
First, we need to find the concentration gradient. We know that the solubility of CO2 at 10 atm is 6.97 x 10^-4 moles/cm^3. Assuming the concentration decreases linearly from the container to the outside (steady-state), we can calculate the concentration gradient:
dC/dx = (C_initial - C_final) / thickness
C_initial is the initial concentration inside the container, which is 6.97 x 10^-4 moles/cm^3.
C_final is the final concentration outside the container, which is assumed to be 0 moles/cm^3 (insignificant amount in the surroundings).
The thickness is given as 100 microns, which is equal to 0.01 cm.
dC/dx = (6.97 x 10^-4 - 0) / 0.01
Now, we can calculate the molar flux:
J = -D * dC/dx
J = - (2.29 x 10^-8 cm^2/s) * [(6.97 x 10^-4 - 0) / 0.01]
Next, we need to find the time it takes for 0.001 moles of CO2 to leak. The molar flux is given in moles/(cm^2.s), so we can find the time by rearranging Fick's Law:
Time = (0.001 moles) / J
Therefore, the time it takes for 0.001 moles of CO2 to leak from the container at steady-state is:
Time = (0.001 moles) / [(2.29 x 10^-8 cm^2/s) * [(6.97 x 10^-4 - 0) / 0.01]]
To convert the time to hours, remember to adjust the units:
1 hour = 3600 seconds
(b) To find the activation energy for diffusion, we can use the Arrhenius equation:
D2 / D1 = exp(-Ea / RT2) / exp(-Ea / RT1)
Where:
- D2 is the diffusivity at the higher temperature
- D1 is the diffusivity at the lower temperature
- Ea is the activation energy (in eV)
- R is the gas constant (8.314 J/(mol.K))
- T2 and T1 are the higher and lower temperatures, respectively
Since we are given the diffusivity doubling when the temperature increases by 10 degrees C, we can write:
(D1 * 2) / D1 = exp(-Ea / RT2) / exp(-Ea / RT1)
Simplifying:
2 = exp(-Ea / R(T2 + 10)) / exp(-Ea / RT1)
Taking the natural logarithm of both sides:
ln(2) = (-Ea / R(T2 + 10)) + (Ea / RT1)
ln(2) * R(T2 + 10) = Ea * (1/T1 - 1/(T2 + 10))
Now we can solve for Ea:
Ea = (ln(2) * R(T2 + 10)) / (1/T1 - 1/(T2 + 10))
Substitute the values of R, T1, and T2 into the equation to find the activation energy in eV.